Question

100 ml of 0.1 M NaOH is titrated against 0.1 M HCL taken in the burette. what is molarity of resulting solution when the alkali is half neutralized.

Answer 1

0.1 L x 0.1 M = 0.01 mol NaOH

half neutralized means 0.01 / 2 = 0.005 mol NaOH left

To neutralize half NaOH 50 mL of HCl needed, so the final volume:

100 + 50 = 150 mL = 0.15 L

Resulting molarity = 0.005 mol / 0.15 L = 0.033 M NaOH